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Q. No. 1 5 Carry One Mark Each
1. Choose the most appropriate phrase from the options given
below to complete the following sentence.
The aircraft_______ take off as soon as its flight plan was
filed. (A) is allowed to (B) will be allowed to (C) was allowed to
(D) has been allowed to Answer: (C)
2. Read the statements: All women are entrepreneurs. Some women
are doctors Which of the following conclusions can be logically
inferred from the above statements? (A) All women are doctors (B)
All doctors are entrepreneurs (C) All entrepreneurs are women (D)
Some entrepreneurs are doctors Answer: (D)
3. Choose the most appropriate word from the options given below
to complete the following sentence.
Many ancient cultures attributed disease to supernatural causes.
However, modern science has largely helped _________ such
notions.
(A) impel (B) dispel (C) propel (D) repel Answer: (B)
4. The statistics of runs scored in a series by four batsmen are
provided in the following table, Who is the most consistent batsman
of these four?
Batsman Average Standard deviation K 31.2 5.21 L 46.0 6.35 M
54.4 6.22 N 17.9 5.90
(A) K (B) L (C) M (D) N Answer: (A) Exp: If the standard
deviation is less, there will be less deviation or batsman is more
consistent
5. What is the next number in the series? 12 35 81 173 357 ____
Answer: 725
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Exp:
357368725
Q. No. 6 10 Carry One Mark Each
6. Find the odd one from the following group: W,E,K,O I,Q,W,A
F,N,T,X N,V,B,D (A) W,E,K,O (B) I,Q,W,A (B) F,N,T,X (D) N,V,B,D
Answer: (D) Exp:
Difference of position: D
7. For submitting tax returns, all resident males with annual
income below Rs 10 lakh should fill up Form P and all resident
females with income below Rs 8 lakh should fill up Form All people
with incomes above Rs 10 lakh should fill up Form R, except non
residents with income above Rs 15 lakhs, who should fill up Form S.
All others should fill Form T. An example of a person who should
fill Form T is
(A) a resident male with annual income Rs 9 lakh (B) a resident
female with annual income Rs 9 lakh
(C) a non-resident male with annual income Rs 16 lakh (D) a
non-resident female with annual income Rs 16 lakh Answer: (B) Exp:
Resident female in between 8 to 10 lakhs havent been mentioned.
8. A train that is 280 metres long, travelling at a uniform
speed, crosses a platform in 60 seconds and passes a man standing
on the platform in 20 seconds. What is the length of the platform
in metres?
Answer: 560 Exp: For a train to cross a person, it takes 20
seconds for its 280m. So, for second 60 seconds. Total distance
travelled should be 840. Including 280 train length
so length of plates =840-280=560
12 35 81 173 357 ________
23 46 92 184 368
difference
W E K O
8 6 4
1 Q W A
8 6 4
F N T X
8 6 4
N V B D
8 6 2
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9. The exports and imports (in crores of Rs.) of a country from
2000 to 2007 are given in the following bar chart. If the trade
deficit is defined as excess of imports over exports, in which year
is the trade deficit 1/5th of the exports?
(A) 2005 (B) 2004 (C) 2007 (D) 2006 Answer: (D)
Exp: imports exports 10 12004,exports 70 7
= =
26 22005,76 720 12006,
100 510 12007,
100 11
=
=
=
10. You are given three coins: one has heads on both faces, the
second has tails on both faces, and the third has a head on one
face and a tail on the other. You choose a coin at random and toss
it, and it comes up heads. The probability that the other face is
tails is
(A) 1/4 (B) 1/3 (C) 1/2 (D) 2/3 Answer: (B)
120110100908070605040302010
0 2000 2001 2002 2003 2004 2005 2006 2007
Exports Im ports
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Q. No. 1 25 Carry One Mark Each
1. For matrices of same dimension M, N and scalar c, which one
of these properties DOES NOT ALWAYS hold?
(A) (MT)T = M (B) (cMT)T = c(M)T (C) (M + N)T = M T + NT (D) MN
= NM Answer: (D) Exp: Matrix multiplication is not commutative in
general.
2. In a housing society, half of the families have a single
child per family, while the remaining half have two children per
family. The probability that a child picked at random, has a
sibling is _____
Answer: 0.667 Exp: Let 1E = one children family 2E = two
children family and A = picking a child then by Bayes theorem,
required probability is
( )21
.x 2E 2P 0.667A 1 x 1 3. .x
2 2 2
= = =
+
(Here x is number of families)
3. C is a closed path in the z-plane given by z 3.= The value of
the integral 2
C
z z 4jz 2 j
+
+
dz is (A) ( )4 1 j2 pi + (B) ( )4 3 j2pi (C) ( )4 3 j2 pi + (D)
( )4 1 j2pi Answer: (C) Exp:
Z 2j= is a singularity lies inside C : Z 3= By Cauchys integral
formula,
[ ] [ ]
22
CZ 2 j
Z Z 4j dz 2 j. Z Z 4jZ 2j
2 j 4 2 j 4 j 4 3 j2=
+ = pi + +
= pi + + = pi +
4. A real (4 4) matrix A satisfies the equation A2 = I, where I
is the (4 4) identity matrix. The positive eigen value of A is
__________.
Answer: 1
Exp:
2 1A I A A if= = is on eigen value of A then 1
is also its eigen value. Since, we
require positive eigen value. 1 = is the only possibility as no
other positive number is self inversed
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5. Let X1, X2, and X3 be independent and identically distributed
random variables with the uniform distribution on [0, 1]. The
probability P{X1 is the largest} is ________
Answer: 0.32-0.34
6. For maximum power transfer between two cascaded sections of
an electrical network, the relationship between the output
impedance Z1 of the first section to the input impedance Z2 of the
second section is
(A) 2 lZ Z= (B) 2 lZ Z= (C) 2 1Z Z= (D) 2 1Z Z= Answer: (C) Exp:
Two cascaded sections
Z1 = Output impedance of first section Z2 = Input impedance of
second section For maximum power transfer, upto 1st section is
*
L 1*
L 2 1
Z Z
Z Z Z
=
=
7. Consider the configuration shown in the figure which is a
portion of a larger electrical network
For R 1= and currents i1 = 2A, i4 = -1A, i5 = -4A, which one of
the following is TRUE? (A) i6 = 5 A (B) 3i 4A= (C) Data is
sufficient to conclude that the supposed currents are impossible
(D) Data is insufficient to identify the current 2 3 6i , i , and i
Answer: (A)
1Z LZ 2ZSection
1Section
2
5i
2i
4i
1i 6i
3iR R
R
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Exp: Given 14
5
i 2Ai 1Ai 4A
=
=
=
KCL at node A, 1 4 22
i i ii 2 1 1A
+ =
= =
1. KCL at node B, 2 5 33
i i ii 1 4 3A
+ =
= =
KCL at node C, ( )
3 6 1
6
i i ii 2 3 5A
+ =
= =
8. When the optical power incident on a photodiode is 10 W and
the responsivity is 0.8A / W, the photocurrent generated ( )in A is
________.
Answer: 8
Exp: ( ) p0
IResponsivity R
P=
p6
8
I0.8
10 10I 8 A
=
=
9. In the figure, assume that the forward voltage drops of the
PN diode D1 and Schottky diode D2 are 0.7 V and 0.3 V,
respectively. If ON denotes conducting state of the diode and OFF
denotes non-conducting state of the diode, then in the circuit,
(A) both D1 and D2 are ON (B) D1 is ON and D2 is OFF (C) both D1
and D2 are OFF (D) D1 is OFF and D2 is ON Answer: (D) Exp: Assume
both the diode ON. Then circuit will be as per figure (2)
( )
2
1 2
D
D D
1 2
10 0.7I 9.3mA1k
0.7 0.3I 20mA20
Now, I I I
10.7 mA Not possibleD is OFF and hense D ON
= =
= =
=
=
1k 20
101D 2D
1K
10V
20
1DI
2DI
I
0.7V 0.3V
1K
10V
20
1D
2D
( )Figure 1
5i
1
1
1
4i3i
2i
6ii1A
B
C
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10. If fixed positive charges are present in the gate oxide of
an n-channel enhancement type MOSFET, it will lead to
(A) a decrease in the threshold voltage (B) channel length
modulation (C) an increase in substrate leakage current (D) an
increase in accumulation capacitance Answer: (A)
11. A good current buffer has (A) low input impedance and low
output impedance (B) low input impedance and high output impedance
(C) high input impedance and low output impedance (D) high input
impedance and high output impedance Answer: (B) Exp: iIdeal current
Buffer has Z 0= 0Z =
12. In the ac equivalent circuit shown in the figure, if ini is
the input current and RF is very large, the type of feedback is
(A) voltage-voltage feedback (B) voltage-current feedback (C)
current-voltage feedback (D) current-current feedback Answer: (B)
Exp: Output sample is voltage and is added at the input or current
It is voltage shunt negative feedback i.e, voltage-current negative
feedback
13. In the low-pass filter shown in the figure, for a cut-off
frequency of 5kHz, the value of R2 ( )in k is ____________.
2R
C10nF1k
1R oViV
+
DRDR
2M
1M
FRsmall signalinput i in
out
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Answer: 3.18 Exp: f 5KHz=
( )2
2 3 9
1Cut off frequency LPF 5KHzR C
1R 3.18k2 5 10 10 10
= =
2pi
= = pi
14. In the following circuit employing pass transistor logic,
all NMOS transistors are identical with a threshold voltage of 1 V.
Ignoring the body-effect, the output voltages at P, Q and R
are,
(A) 4 V, 3 V, 2 V (B) 5 V, 5 V, 5 V (C) 4 V, 4 V, 4 V (D) 5 V, 4
V, 3 V
Answer: (C) Exp: Assume al NMOS are in saturation
( )
( ) ( )( ) ( )
( )( ) ( )
1
1
DS GS T
1
p p
p p
2D GS T
2D p
V V VFor m
5 V 5 V 1
5 V 4 V Sat
I k V V
I K 4 V ........ 1
>
=
=
( )( ) ( )
( ) ( )
1
2
1 2
22
D Q2
D Q
D D
2 2p Q
p Q p Q
p Q
For m ,
I K 5 V 1
I K 4 V ...... 2
I I
4 V 4 V
V V & V V 8V V 4V
=
=
=
=
= + =
= =
( )
( ) ( )
3
2 3
32
D R
D D
2 2Q R
R Q
p Q R
For m ,
I K 5 V 1I I
4 V 4 V
V V 4VV V V 4V
=
=
=
= =
= = =
5V 5V 5V
RQP
5V
5V
1M
2M
3M
P
Q
R
5V
5V
5V
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15. The Boolean expression ( ) ( ) ( )X Y X Y X Y X+ + + + +
simplifies to (A) X (B) Y (C) XY (D) X+Y Answer: (A) Exp: Given
Boolean Expression is ( )( )X Y X Y XY X+ + + + As per the
transposition theorem
( ) ( )( )( ) ( )
( )( ) ( )( )
( )
A BC A B A C
so, X Y X Y X YY X 0
X Y X Y XY X X XY .X
X X Y .X X XX. Y.X X 0 Y.X
Applyabsorption theorem X 1 Y X.1 X
+ = + +
+ + = + = +
+ + + + = +
= + + = + + = + +
= + = =
16. Five JK flip-flops are cascaded to form the circuit shown in
Figure. Clock pulses at a frequency of 1 MHz are applied as shown.
The frequency (in kHz) of the waveform at Q3 is __________ .
Answer: 62.5 Exp: Given circuit is a Ripple (Asynchrnous)
counter. In Ripple counter, o/p frequency of each
flip-flop is half of the input frequency if their all the states
are used otherwise o/p frequency
of the counter is input frequencymodulus of the counter
=
So, the frequency at 36
z
input frequencyQ16
1 10 H 62.5kHz16
=
= =
17. A discrete-time signal [ ] ( )2x n sin n ,n beingan
integer,is= pi (A) periodic with period pi . (B) periodic with
period 2pi . (C) periodic with period / 2pi . (D) not periodic
Answer: (D)
1
1
J4 Q4clk>
K4
1
1
J3 Q3K2
J2 Q2clk>
K2
1
1
1
1
J1 Q1clk>
K1clk>
J0
K0
1
1clk>
clock
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Exp: Assume [ ]x n to be periodic, (with period N)
[ ] [ ]( ) ( )( )2 2
x n x n N
sin n sin n N
= +
pi = pi +
Every frigonometric function repeate after 2pi interval.
( ) ( )2 2 22
sin n 2 k sin h N
2k2 k N N
pi + pi = pi + pi
pi = pi =
pi
Since k is any integer, there is no possible value of k for
which N can be an integer, thus non-periodic.
18. Consider two real valued signals, x(t) band-limited to [
]500 Hz, 500Hz and ( )y t band-limited to [ ]1kHz, 1kHz . For ( ) (
) ( )z t x t . y t ,= the Nyquist sampling frequency (in kHz) is
__________
Answer: 3 Exp: ( )x t is band limited to [ ]500Hz, 500Hz ( ) [
]y t is band limited to 1000Hz, 1000Hz ( ) ( ) ( )z t x t .y t=
Multiplication in time domain results convolution in frequency
domain. The range of convolution in frequency domain is [ ]1500Hz,
1500Hz So maximum frequency present in z(t) is 1500Hz Nyquist rate
is 3000Hz or 3 kHz
19. A continuous, linear time-invariant filter has an impulse
response h(t) described by
( ) {3 for 0 t 30 otherwiseh t = When a constant input of value
5 is applied to this filter, the steady state output is _______.
Answer: 45 Exp:
( ) ( ) ( )y t x t * h t= ( )x t =
( )h t =
( ) ( )3
0
y t 3.5.d 45 steady state output= =
( )x t ( )y t( )h t
5
t
3
3t
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20. The forward path transfer function of a unity negative
feedback system is given by
( ) ( ) ( )KG s
s 2 s 1=
+
The value of K which will place both the poles of the
closed-loop system at the same location, is ______.
Answer: 2.25
Exp: Given ( ) ( )( )( )
KG ss 2 s 1
H s 1
=
+
=
Characteristic equation: ( ) ( )
( )( )
1 G s H s 0K1 0
s 2 s 1
+ =
+ =+
The poles are 1,29
s 1 4K4
=
If 9 K 0,4
= then both poles of the closed loop system at the same
location.
So, 9K 2.254
=
21. Consider the feedback system shown in the figure. The
Nyquist plot of G(s) is also shown. Which one of the following
conclusions is correct?
(A) G(s) is an all-pass filter (B) G(s) is a strictly proper
transfer function (C) G(s) is a stable and minimum-phase transfer
function (D) The closed-loop system is unstable for sufficiently
large and positive k Answer: ( D) Exp: For larger values of K, it
will encircle the critical point (-1+j0), which makes
closed-loop
system unstable.
( )ImG j
1+ ( )ReG j1+
k ( )G s
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22. In a code-division multiple access (CDMA) system with N = 8
chips, the maximum number of users who can be assigned mutually
orthogonal signature sequences is ________
Answer: 7.99 to 8.01
Exp: Spreading factor(SF)= chip ratesymbol rate
This if a single symbol is represented by a code of 8 chips Chip
rate =80symbol rate
S.F (Spreading Factor) 8 symbol rate 8symbol rate
= =
Spread factor (or) process gain and determine to a certain
extent the upper limit of the total number of uses supported
simultaneously by a station.
23. The capacity of a Binary Symmetric Channel (BSC) with
cross-over probability 0.5 is ________
Answer: 0 Exp: Capacity of channel is 1-H(p) H(p) is entropy
function With cross over probability of 0.5
( ) 2 21 1 1 1H p log log 12 0.5 2 0.5Capacity 1 1 0
= + =
= =
24. A two-port network has sattering parameters given by [ ] 11
1221 22
S SS
S S
=
. If the port-2 of the
two-port is short circuited, the 11S parameter for the resultant
one-port network is
( ) 11 11 22 12 2122
s s s s sA1 s
+
+ ( ) 11 11 22 12 21
22
s s s s sB1 s
+
( ) 11 11 22 12 2122
s s s s sC1 s
+
( ) 11 11 22 12 2122
s s s s sD1 s
+
Answer:(B) Exp:
1 11 1 12 2
2 21 1 22 2
b s a s ab s a s a
= +
= +
1 11 12 1
2 21 22 2
b s s ab s s a
=
; 2
11
1 a 0
bs
a=
=
By verification Answer B satisfies.
Two portNetwork
1a2a
1b2b
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25. The force on a point charge +q kept at a distance d from the
surface of an infinite grounded metal plate in a medium of
permittivity is
(A) 0 (B) 2
2q
away from the plate16 dpi
(C) 2
2q
towards the plate16 dpi
(D) 2
2q
towards the plate4 dpi
Answer:(C)
Exp:
( )
1 22
2 2
2 2
Q Q1F4 R
1 9 9F4 16 d2d
=
pi
= =
pi pi
Since the charges are opposite polarity the force between them
is attractive.
Q.No. 26 55 Carry Two Marks Each
26. The Taylor series expansion of 3 sin x + 2 cos x is
( ) 32 xA 2 3x x .......2
+ + ( )3
2 xB 2 3x x .......2
+ +
( ) 32 xC 2 3x x .......2
+ + + + ( ) 32 xD 2 3x x .......2
+ +
Answer: (A)
Exp:
3 2x x3sin x 2cos x 3 x ... 2 1 ...3! 2!
+ = + + +
32 x2 3x x ...
2= + +
27. For a Function g(t), it is given that ( ) 2j t 2g t e dt
e+
= for any real value . If
( ) ( ) ( )ty t g d , then y t dt+
= is
(A)0 (B)-j (C) - j2
(D) j2
Answer: (B) Exp: Given
( ) ( )( )( )
2jwt 2wg t .e dt .e let G j
g t dt 0
=
=
d
d
q
q+
metal plate
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( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )
( ) ( )
( ) ( ) ( )2
t
j t
2w
y t g z .dz y t g t * u t u t in unit step function
Y j G j .U j
Y j y t .e dt
1Y j0 y t dt .e 0j1 jj
= =
=
=
= = + pi =
= =
28. The volume under the surface z(x, y) = x + y and above the
triangle in the x-y plane defined by { }0 y x and 0 x 12
is___________.
Answer: 864
Exp:
( ) ( )12 x
R x 0 y 0
Volume Z x, y dydx x y dydx= =
= = +
x 1212 122 32
x 0 00 0
y 3 3 xxy .dx x dx 864
2 2 2 3=
= + = = =
29. Consider the matrix:
6
0 0 0 0 0 10 0 0 0 1 00 0 0 1 0 0
J0 0 1 0 0 00 1 0 0 0 01 0 0 0 0 0
=
Which is obtained by reversing the order of the columns of the
identity matrix 6I . Let 6 6P I J ,= + where is a non-negative real
number. The value of for which det(P) =
0 is ___________. Answer: 1
Exp: ( ) 2 2Consider, i Let P I J= + 1 0 0 1 10 1 1 0 1
= + =
( )
2
4 4
P 1
1 0 0
ii Let P I J
=
0 1 0
= + = 0 1 0 0 0 1
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( ) ( )
( ) ( ) ( ) ( )
( )
22 2 2
6 6
32
1 0 0 1P 1 1 0 0 1
0 0 1 0 0
1 1 1
S im ilarly , if P I J then w e get
P 1
P 0 1, 1
is non negative
1
=
= =
= +
=
= =
=
30. A Y-network has resistances of 10 each in two of its arms,
while the third arm has a resistance of 11 in the equivalent
network, the lowest value ( )in among the three resistances is
______________.
Answer: 29.09 Exp:
X 29.09y 32z 32
= = =
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
10 10 10 11 10 11X
1110 10 10 11 10 11
y10
10 10 10 11 10 11z
10
+ +=
+ +=
+ +=
i.e, lowest value among three resistances is 29.09
31. A 230 V rms source supplies power to two loads connected in
parallel. The first load draws 10 kW at 0.8 leading power factor
and the second one draws 10 kVA at 0.8 lagging power factor. The
complex power delivered by the source is
(A) (18 + j 1.5) kVA (B) (18 - j 1.5) kVA (C) (20 + j 1.5) kVA
(D) (20 - j 1.5) kVA
Star Connection
10
1011
X
Y
Z10
10 11
Delta Connection
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Answer: (B) Exp:
Load 1:
I
P 10kwcos 0.8 S P jQ 10 j7.5 KVAQ P tan 7.5KVAR
= = = =
= =
Load 2: S 10 KVA=
Qcos 0.8 sin
S
Pcos
S
P0.8 P 8kw Q 6KVAR10
= =
=
= = =
IS P jQ 8 j6= + = + Complex power delivered by the source is I
IIS S 18 j1.5 KVA+ =
32. A periodic variable x is shown in the figure as a function
of time. The root-mean-square (rms) value of x is_______.
Answer: 0.408
x
1
0
T / 2 T / 2t
LoadI
LoadII
+
230V
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Exp:
( )( )T 2rms0
1x x t dt
T=
( )
( )2T 2 T
2
T0 2T
3 2
203
rms 3
2 Tt 0 t 2Tx tT0 t T2
21.t .dt 0 .dt
T T
1 4 t.
T T 3
4 T 1x . 0.408
3T 8 6
=
= +
=
=
33. In the circuit shown in the figure, the value of capacitor
C(in mF) needed to have critically damped response i(t)
is____________.
Answer: 10mF Exp: By KVL,
( ) ( ) ( ) ( )di t 1v t Ri t L. i t dtdt C
= + +
Differentiate with respect to time,
( ) ( ) ( )( ) ( ) ( )
2
2
2
2
1,2
2
1,2
R.di t di ti i tR0 . 0dt L dt LC
d i t di t i tR. 0
dt L dt LCR R 4L L LCD
2R R 1D
2L 2L LC
= + + =
+ + =
=
=
For critically damped response,
2
2R 1 4LC F2L LC R
= =
Given, L=4H; R= 40
( )24 4C 10mF40
=
40 4 H C
( )i t+
OV
X
1
( )0,0 T2 Tt
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34. A BJT is biased in forward active mode, Assume BEV 0.7V,kT /
q 25mV= = and reverse saturation current 13SI 10
= A. The transconductance of the BJT (in mA/V) is ________.
Answer: 5.785
Exp: 13BE sKTV 0.7V, 25mV, I 10q
= = =
BE T
Cm
T
V /VC S
13 0.7/25mV
Cm
T
ITransconductance, gV
I I e 1
10 e 1 144.625mAI 144.625mAg 5.785A / VV 25mV
=
=
= =
= = =
35. The doping concentrations on the p-side and n-side of a
silicon diode are 16 31 10 cm and 17 31 10 cm , respectively. A
forward bias of 0.3 V is applied to the diode. At T = 300K, the
intrinsic carrier concentration of silicon 10 3in 1.5 10 cm
= and kT 26mV.q
= The electron
concentration at the edge of the depletion region on the p-side
is (A) 9 32.3 10 cm (B) 16 31 10 cm (C) 17 31 10 cm (D) 6 32.25 10
cm Answer:(A)
Exp: bi T2
V /Vi
A
nElectron concentration, n eN
( )210 0.3/26mV16
9 3
1.5 10e
1 102.3 10 / cm
=
=
36. A depletion type N-channel MOSFET is biased in its linear
region for use as a voltage controlled resistor. Assume threshold
voltage
TH GS DSV 0.5V,V 2.0V, V 5V,W / L 100,= = = = 8 2
OXC 10 F / cm
= and 2n 800cm / V s = . The value of the resistance of the
voltage controlled resistor ( )in is ________.
Answer:500
Exp: x
8T GS DS
WGiven V 0.5V; V 2V; V 5V; 100; C 10 f / cmL
= = = = =
( )
( )
2n
2D n 0x GS T DS DS
1 12D
ds n 0x GS T DS DSDS DS
800cm / v s1 WI C 2 V V V V2 L
I 1 Wr C 2 V V V V
V V 2 L
=
=
=
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EC-GATE-2014 PAPER-01|
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( )1
n 0x GS T n 0x DSW WC V V C VL L
=
( )
( )
ds
n 0x GS T Ds
8
1r WC V V V
L1 500
800 10 100 2 0.5 5
=
= = +
37. In the voltage regulator circuit shown in the figure, the
op-amp is ideal. The BJT has BEV 0.7V= and 100, = and the zener
voltage is 4.7V. For a regulated output of 9 V, the
value of ( )R in is ______ .
Answer:1093 Exp: BE Z 0Given V 0.7V, 100, V 4.7V, V 9V= = =
=
( )R
R z
RV 9R 1k
R4.7 9 V VR 1k
R 1093
= +
= =+
=
38. In the circuit shown, the op-amp has finite input impedance,
infinite voltage gain and zero input offset voltage. The output
voltage outV is
(A) ( )2 1 2I R R + (B) 2 2I R
(C) 1 2I R
(D) ( )1 1 2I R R +
2R
1R1l
2l
+outV
IV 12V= 0V 9 V=
1k
zV 4.7 V=
+
1k
R
=i
V 12V 9V
1K
RV
R
+
zV
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EC-GATE-2014 PAPER-01|
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Answer: (C) Exp: iGiven, Z =
( )( )
L
0
0
i
2 1 2 1
1 21
1 2
AV 0
V R / /R IR R I ...... 1
R R
=
=
=
=
+
KCL at inverting node
( )2 02 i1 2
02
2 1 2
0 1 2 2 11
2 1 2 1 2
0 1 2
V VV 0 ZR R
V 1 1VR R R
V R R R RIR R R R R
V I R
+ = =
= +
+=
+
=
39. For the amplifier shown in the figure, the BJT parameters
are BEV 0.7V, 200,= = and thermal voltage TV 25mV.= The voltage
gain ( )0 iv / v of the amplifier is _______.
Answer: -237.76 Exp: BE TV 0.7V, 200, V 25mV= = = DC
Analysis:
B
E
E
11kV 12 3V11k 33k
V 3 0.7 2.3V2.3I 2.277mA
10 1k
= =+
= =
= =
+
1R
2R
0V2V
1V
+
1I
CCV 12V= +
CR5k1R
33k1 F
iv
2R11k
1 Fov
SR10
1ER
1k EC
1mF
-
EC-GATE-2014 PAPER-01|
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( )( ) ( )
B
C
e
0 CV
i e s
V
I 11.34 AI 2.26mA
25mVr 10.98
2.277mAV R 200 5kAV r 1 R 200 10.98 201 10
A 237.76
= =
= =
= = = + + +=
40. The output F in the digital logic circuit shown in the
figure is
( )A F XYZ XYZ= + ( )B F XYZ XYZ= + ( )C F XYZ XYZ= + ( )D F XYZ
XYZ= + Answer: (A) Exp:
Assume dummy variable K as a output of XOR gate K X Y XY XY= =
+
( )( )
( )
F K. K Z
KZ K.Z
K. KZ K.K.Z
0 K.Z K. K 0 and K.K K
=
= +
= +
= + = =
Put the value of K in above expression
( )F XY XY ZXYZ XYZ
= +
= +
XORX
Y AND
F
Z
XNOR
X
Y
Z
XOR
XNOR
K
F
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EC-GATE-2014 PAPER-01|
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41. Consider the Boolean function, ( )F w,x,y,z wy xy wxyz wxy
xz xyz.= + + + + + which one of the following is the complete set
of essential prime implicants?
(A) w,y, xz, x z (B) w,y,xz (C) y,x yz (D) y,xz,xz Answer: (D)
Exp: Given Boolean Function is ( )F w,x, y,z wy xy wxyz wxy xz xyz=
+ + + + + By using K-map
So, the essential prime implicants (EPI ) are y, xz, xz
42. The digital logic shown in the figure satisfies the given
state diagram when Q1 is connected to input A of the XOR gate.
Suppose the XOR gate is replaced by an XNOR gate. Which one of
the following options preserves the state diagram?
(A) Input A is connected to Q2 (B) Input A is connected to Q2
(C) Input A is connected to Q1 and S is complemented (D) Input A is
connected to Q1 Answer: (D)
D1 Q1
Q1CLK>
A
S
D2 Q2
Q2>
S 0=
S 1=00 01
10 11
S 1=S 0=
S 0=
S 1=
S 0=
S 1=
00 01 11 10
00 1 1 1
01 1 1 1
11 1 1 1
10 1 1 1
x z
xz
y
wxyz
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EC-GATE-2014 PAPER-01|
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Exp: The input of D2 flip-flop is ( )2 1 1 1D Q s Q s A Q= + =
The alternate expression for EX-NOR gate is A B A B A B= = =
So, if the Ex-OR gate is substituted by Ex-NOR gate then input A
should be connected to 1Q
( )2 1 1 1 1 1i 1
D Q S Q S Q S Q .S A QQ S Q .S
= + = + =
= +
43. Lex [ ] ( ) ( )n n1 1
x n u n u n 19 3
=
. The Region of Convergence (ROC) of the z-
transform of x[n]
(A) 1is z9
> (B) 1is z3
< (C) 1 1is z3 9
> > (D) does not exist.
Answer: (C)
Exp: [ ] [ ] [ ]n n1 1Given x n u n u n 1
9 3
=
[ ]h
oc
1 1for u n R in z9 9
>
(Right sided sequence, ocR in exterior of circle of radius 19
)
Thus overall oc1 1R in z9 3
< <
44. Consider a discrete time periodic signal x[n] = nsins
pi
. Let ka be the complex Fourier
series coefficients of x[n]. The coefficients { }ka are non-zero
when k = Bm 1, where m is any integer. The value of B
is_________.
Answer: 10
Exp: [ ] nGiven x n sin ; N 105pi
= =
Fourier series co-efficients are also periodic with period N
10=
[ ]2 2j n i n10 101 1x n e e
2j 2 jpi pi
=
1 1 1 1 10 9
1 1 1 1
1 11 1
1 1 1a ; a a a a
2j 2 j 2 ja a 10 a a 20
ora a 20a a 10
k 10m 1 or k 10.m 1 B 10
+
= = = = =
= + = +
= += +
= + = =
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EC-GATE-2014 PAPER-01|
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45. A system is described by the following differential
equation, where u(t) is the input to the system and y(t) is the
output of the system.
( ) ( ) ( ).y t 5y t u t+ = When y(0) = 1 and u(t) is a unit
step function, y(t) is (A) 5t0.2 0.8e+ (B) 5t0.2 0.2e (C) 5t0.8
0.2e+ (D) 5t0.8 0.8e Answer: (A) Exp: Given ( ) ( ) ( ) ( ) ( )y t
5y t u t and y 0 1; u t isa unit stepfunction.+ = = Apply Laplace
transform to the given differential equation.
( ) ( ) ( )
( )[ ] ( ) ( ) ( ) ( )
( ) ( )( ) ( )( )
( ) ( )
1S y s y 0 5y ss
1 dy 1y s s 5 y 0 L s y s y 0 L u tss dt
1 1sy ss 5s 1 A By s
s s 5 s s 51 4A ; B5 5
1 4y s5s 5 s 5
+ =
+ = + = =
+=
+
+= +
+ +
= =
= ++
Apply inverse Laplace transform,
( )( )
5t
5t
1 4y t e5 5
y t 0.2 0.8e
= +
= +
46. Consider the state space model of a system, as given
below
[ ]
.
11 1
.
2 2 2.
3 33
x 1 1 0 x 0 xx 0 1 0 x 4 u; y 1 1 1 x
0 0 2 x 0 xx
= + =
The system is (A) controllable and observable (B) uncontrollable
and observable (C) uncontrollable and unobservable (D) controllable
and unobservable Answer: (B)
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EC-GATE-2014 PAPER-01|
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Exp: From the given state model,
[ ]
2c
c
c c
02
0
0
01 1 0A 0 1 0 B 4 c 1 1 1
0 0 2 0
Controllable: Q c B AB A Bif Q 0 controllable
0 4 8Q 4 4 4 Q 0
0 0 0
uncontrollable
CObservable : Q CA
CA
If Q 0 observable1 1 1
Q 1 0 21 1 4
= = =
= =
= =
=
=
0Q 1
Observable
=
The system is uncontrollable and observable
47. The phase margin in degrees of ( ) ( ) ( ) ( )10G s
s 0.1 s 1 s 10=
+ + + + calculated using the
asymptotic Bode plot is_______. Answer: 48
Exp:
( ) ( )( )( )( )
[ ]( ) [ ][ ][ ]
10G ss 0.1 s 1 s 10
10G ss s0.1 1 1 s 1 .10
0.1 1010G s
1 10s 1 s 1 0.1s
=
+ + +
=
+ + +
=
+ + +
By Approximation, ( ) [ ]10G s
10s 1=
+
-
EC-GATE-2014 PAPER-01|
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Phase Margin = gc
1
180 GH
10 0.99180 tan1
Phase Margin =95 .73
=
= +
=
gc 2
2
2gc
101100 1
99100199 0.9949r / sc
1
= = +
= =
=
Asymptotic approximation, Phase margin = 45 48
48. For the following feedback system ( ) ( ) ( )1G s
s 1 s 2=
+ + +. The 2% settling time of the step
response is required to be less than 2 seconds.
Which one of the following compensators C(s) achieves this?
( ) 1A 3s 5 +
( ) 0.03B 5 1s
+
( ) ( )C 2 s 4+ ( ) s 8D 4s 3
+ +
Answer: (C) Exp: By observing the options, if we place other
options, characteristic equation will have 3rd order
one, where we cannot describe the settling time. ( ) ( )If C s 2
s 4= + is considered The characteristic equation, is
2
2
s 3s 2 2s 8 0s 5s 10 0+ + + + =
+ + =
Standard character equation 2 2n n2n n
s 2 s 0
10; 2.5+ + =
= =
Given, 2% settling time, n
n
4 2 w 2w
< >
49. Let x be a real-valued random variable with E[X] and E[X2]
denoting the mean values of X and X2, respectively. The relation
which always holds true is
( ) [ ]( )2 2A E X E X > ( ) [ ]( )22B E X E X ( ) [ ]( )22C
E X E X = ( ) [ ]( )22D E X E X > Answer: (B) Exp: ( ) ( ) ( ){
}22V x E x E x 0 i.e., var iance cannot be negative=
( ) ( ){ }22E x E x
r + ( )C s ( )G s y
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EC-GATE-2014 PAPER-01|
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50. Consider a random process ( ) ( )X t 2 sin 2 t ,= pi + where
the random phase is uniformly distributed in the interval [ ]0,2pi
. The auto-correlation ( ) ( )1 2E X t X t
( ) ( )( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )
1 2 1 2
1 2 1 2
A cos 2 t t B sin 2 t t
C sin 2 t t D cos 2 t t
pi + pi
pi + pi
Answer: (D) Exp: Given ( )X(t) 2 sin 2 t= pi + in uniformly
distributed in the interval [ ]0,2pi
[ ] ( )21 2 1 20E x(t )x(t ) 2 sin(2 t ) 2 sin 2 t f ( )dpi
= pi + pi +
( ) ( )2 1 202 2
1 2 1 20 0
12 sin 2 t sin 2 t . .d2
1 1sin(2 (t t ) 2 )d cos(2 (t t )d
2 2
pi
pi pi
= pi + pi + pi
= pi + + + pi pi pi
First integral will result into zero as we are integrating from
0 to 2 .pi Second integral result into { }1 2cos 2 (t t )pi [ ] (
)1 2 1 2E X(t )X(t ) cos 2 (t t = pi
51. Let ( )Q be the BER of a BPSK system over an AWGN channel
with two-sided noise power spectral density N0/2. The parameter is
a function of bit energy and noise power spectral density.
A system with tow independent and identical AWGN channels with
noise power spectral density N0/2 is shown in the figure. The BPSK
demodulator receives the sum of outputs of both the channels.
If the BER of this system is ( )Q b , then the value of b is
_____________. Answer: 1.414
Exp: O
2E EBit error rate for BPSK Q . Q NNO2
=
O
2EYN
=
0 /1 BPSKModulator
AWGNChannel1
AWGNChannel 2
BPSKDemodulator
0 /1+
12pi
f ( )
0 2pi
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EC-GATE-2014 PAPER-01|
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Function of bit energy and noise OSDNP2
Counterllation diagram of BPSK Channel is WGNA which implies
noise sample as independent
1 11 2
1
11 2
1
1O
1 1
1 1O
Let 2x n n x n
where x 2x
n n n
2ENow Bit error rate QN
E is energy in x
N is PSD of h
+ + = +
=
= +
=
1E 4E= [as amplitudes are getting doubled]
1O ON N= [independent and identical channel]
O O
4E 2EBit error rate Q Q 2 b 2 or 1.414N N
= = =
52. A fair coin is tossed repeatedly until a Head appears for
the first time. Let L be the number of tosses to get this first
Head. The entropy H(L) in bits is _________.
Answer: 2 Exp: In this problem random variable is L
{ }
{ }
{ }
{ } 2 2 2
L can be 1,2,..............
1P L 12
1P L 24
1P L 38
1 1 1 1 1 1 1 1 1H L log lgo log ......... 0 1. 2. 3. .........1
1 12 4 8 2 4 82 4 8
= =
= =
= =
= + + + = + + + +
[ Arithmatic gemometric series summation]
2
1.12 2 211 12 12
= + =
( )2 t
a a ( )1 t
+
+
+x
+1
x n
+2
x n
noise in channel 1
+ +1 2
2x n n
noise in channel 2
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EC-GATE-2014 PAPER-01|
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53. In spherical coordinates, let a .a denote until vectors
along the , directions.
( )100 E sin cos t r a V / mr
= and
( )0.265 H sin cos t r a A / mr
= represent the electric and magnetic field components of the EM
wave of large distances r
from a dipole antenna, in free space. The average power (W)
crossing the hemispherical shell located at r 1km,0 / 2is _______=
pi
Answer: 55.5
Exp:
J r100E sin er
=
( )
( )
( )( )
J rQ
*
avg Qs
2 22s
2avt
s
223
0 Q 0
0.265H sin er
1P E H .ds2
100 0.2651sin r sin d d
2 r1P 26.5 sin d d2
213.25 sin d d 13.25. 23
P 55.5w
pipi
= =
=
=
=
=
= = pi
=
54. For a parallel plate transmission line, let v be the speed
of propagation and Z be the characteristic impedance. Neglecting
fringe effects, a reduction of the spacing between the plates by a
factor of two results in
(A) halving of v and no change in Z (B) no change in v and
halving of Z (C) no change in both v and Z (D) halving of both v
and Z Answer: (B)
Exp:
o
r
276 dZ logr
=
d distance between the two plates so, zo changes, if the spacing
between the plates changes.
1VLC
= independent of spacing between the plates
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EC-GATE-2014 PAPER-01|
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55. The input impedance of a 8
section of a lossless transmission line of characteristic
impedance 50 is found to be real when the other end is
terminated by a load ( )LZ R jX .= + if X is 30 , the value of R (
)in is _________
Answer: 40 Exp:
( )o
L oin o
o L
L L Lin
L L L
Given,s
Z 50Z JZZ Z8 Z KZ
Z J50 Z J50 50 JZZ 50 5050 JZ 50 JZ 50 JZ
=
= +
= = +
+ + = = + +
( )
( )
2 2L L L
in 2 2L
in
mg in
2 2L
2 2L2 2 2
2 2 2 2 2
50Z 50Z J 50 ZZ 50
50 Z
Given , Z RealSo, I Z 0
50 Z 0Z 50R X 50R 50 X 50 30R 40
+ + =
+
=
=
=
+ =
= =
=